Question 1 |
The following differential equation has
3(\frac{d^{2}y}{dt^{2}})+4(\frac{dy}{dt})^{3}+y^{2}+2=x
3(\frac{d^{2}y}{dt^{2}})+4(\frac{dy}{dt})^{3}+y^{2}+2=x
degree = 2, order = 1 | |
degree = 3, order = 2 | |
degree = 4, order = 3 | |
degree = 2, order = 3 |
Question 1 Explanation:
Order is highest derivative term. Degree is power
of highest derivative term.
Question 2 |
Choose the function f (t); -\infty \lt t \lt \infty for which a Fourier series cannot be defined.
3sin(25t) | |
4cos(20t+3)+2sin(10t) | |
exp(-|t|)sin(25t) | |
1 |
Question 2 Explanation:
All other functions are either periodic or constant
function.
Question 3 |
A fair dice is rolled twice. The probability that an odd number will follow an
even number is
1/2 | |
1/6 | |
1/3 | |
1/4 |
Question 3 Explanation:
\begin{aligned} P_{0}=\frac{3}{6}=\frac{1}{2} \\ P_{e}=\frac{3}{6}=\frac{1}{2} \end{aligned}
since both events are independent of each other.
P_{\text {(odd/even) }}=\frac{1}{2} \times \frac{1}{2}=\frac{1}{4}
since both events are independent of each other.
P_{\text {(odd/even) }}=\frac{1}{2} \times \frac{1}{2}=\frac{1}{4}
Question 4 |
A solution of the following differential equation is given by \frac{d^{2}y}{dx^{2}}-5\frac{dy}{dx}+6y=0
y=e^{2x}+e^{-3x} | |
y=e^{2x}+e^{3x} | |
y=e^{-2x}+e^{3x} | |
y=e^{-2x}+e^{-3x} |
Question 4 Explanation:
\begin{aligned} A E \Rightarrow\quad D^{2}-5 D+6&=0 \\ (D-2)(D-3) &=0 \\ D &=2,3 \\ \therefore \quad y &=e^{2 x}+e^{3 x} \end{aligned}
Question 5 |
The function x(t) is shown in the figure. Even and odd parts of a unit step
function u(t) are respectively,
\frac{1}{2},\frac{1}{2}x(t) | |
-\frac{1}{2},\frac{1}{2}x(t) | |
\frac{1}{2},-\frac{1}{2}x(t) | |
-\frac{1}{2},-\frac{1}{2}x(t) |
Question 5 Explanation:
\begin{array}{l} \text { Even part }=\frac{u(t)+u(-t)}{2}\\ \text { Odd part }=\frac{u(t)-u(-t)}{2} \end{array}
Question 6 |
The region of convergence of z-transform of the sequence (\frac{5}{6})^{n}u(n)-(\frac{6}{5})^{n}u(-n-1) must be
|z| \lt \frac{5}{6} | |
|z| \gt \frac{5}{6} | |
\frac{5}{6} \lt |z| \lt \frac{6}{5} | |
\frac{5}{6} \lt |z| \lt \infty |
Question 6 Explanation:
\begin{array}{l} z \text {-transform of }\left(\frac{5}{6}\right)^{n} u(n)-\left(\frac{6}{5}\right)^{n} u(-n-1)\\ =\frac{1}{1-\frac{5}{6} z^{-1}}+\left[1-\frac{1}{1-\left(\frac{6}{5}\right)^{-1} z}\right] \\ \begin{array}{cc} (\mathrm{ROC}) &\quad(\mathrm{ROC})\\ =|z| \gt
\frac{5}{6}&\quad |z| \lt \frac{6}{5} \\ \end{array}\\ \frac{5}{6}\lt |z| \lt \frac{6}{5} \end{array}
Question 7 |
The condition on R, L and C such that the step response y(t) in the figure has
no oscillations, is
R\geq \frac{1}{2}\sqrt{\frac{L}{C}} | |
R\geq \sqrt{\frac{L}{C}} | |
R\geq 2\sqrt{\frac{L}{C}} | |
R=\frac{1}{\sqrt{LC}} |
Question 7 Explanation:
Transfer function
\begin{aligned} =& \frac{\frac{1}{s C}}{R+s L+\frac{1}{s C}}=\frac{1}{s^{2} L C+s C R+1} \\ \frac{Y(s)}{U(s)}=& \frac{\frac{1}{L C}}{s^{2}+\frac{R}{L} s+\frac{1}{L C}} \\ 2 \xi \omega_{n} &=\frac{R}{L} \\ \omega_{n} &=\frac{1}{\sqrt{L C}} \\ \xi &=\frac{R}{2 L} \sqrt{L C}=\frac{R}{2} \sqrt{\frac{C}{L}} \end{aligned}
For no oscillations, \xi \geq 1
\frac{R}{2} \sqrt{\frac{C}{L}} \geq 1 \quad ; \quad R \geq 2 \sqrt{\frac{L}{C}}
\begin{aligned} =& \frac{\frac{1}{s C}}{R+s L+\frac{1}{s C}}=\frac{1}{s^{2} L C+s C R+1} \\ \frac{Y(s)}{U(s)}=& \frac{\frac{1}{L C}}{s^{2}+\frac{R}{L} s+\frac{1}{L C}} \\ 2 \xi \omega_{n} &=\frac{R}{L} \\ \omega_{n} &=\frac{1}{\sqrt{L C}} \\ \xi &=\frac{R}{2 L} \sqrt{L C}=\frac{R}{2} \sqrt{\frac{C}{L}} \end{aligned}
For no oscillations, \xi \geq 1
\frac{R}{2} \sqrt{\frac{C}{L}} \geq 1 \quad ; \quad R \geq 2 \sqrt{\frac{L}{C}}
Question 8 |
The ABCD parameters of an ideal n:1 transformer shown in the figure are \begin{bmatrix} n & 0\\ 0&x \end{bmatrix}
The value of x will be
The value of x will be
n | |
\frac{1}{n} | |
n^{2} | |
\frac{1}{n^{2}} |
Question 8 Explanation:
\begin{aligned} \left[\begin{array}{l} V_{1} \\ I_{1} \end{array}\right] &=\left[\begin{array}{ll} A & B \\ C & D \end{array}\right]\left[\begin{array}{l} V_{2} \\ -I_{2} \end{array}\right] \\ -\frac{I_{2}}{I_{1}} &=\frac{V_{1}}{V_{2}}=\frac{n}{1} \\ V_{1} &=A V_{2}-B I_{2}\\ I_{1}&=C V_{2}-D I_{2} \\ A&=\left.\frac{V_{1}}{V_{2}}\right|_{I_{2}=0}=n \\ D&=-\left.\frac{I_{1}}{I_{2}}\right|_{V_{2}=0}=\frac{V_{2}}{V_{1}}=\frac{1}{n} \end{aligned}
Question 9 |
In a series RLC circuit, R=2k\Omega, L=1H, and C=\frac{1}{400}\mu F The resonant frequency is
2 \times 10^{4}Hz | |
\frac{1}{\pi }\times 10^{4}Hz | |
10^{4}Hz | |
2\pi \times 10^{4}Hz |
Question 9 Explanation:
\begin{aligned} f_{0} &=\frac{1}{2 \pi \sqrt{L C}}=\frac{1}{2 \pi \sqrt{1 \times \frac{1}{400} \times 10^{-6}}} \\ &=\frac{10^{3} \times 20}{2 \pi}=\frac{10^{4}}{\pi} \mathrm{Hz} \end{aligned}
Question 10 |
The maximum power that can be transferred to the load resistor R_{L} from the voltage source in the figure is
1 W | |
10 W | |
0.25 W | |
0.5 W |
Question 10 Explanation:
For maximum power transfer
\begin{aligned} R_{L}&= R_{s}=100 \Omega \\ \therefore \quad P_{\max }&=\frac{V^{2}}{R}=\frac{5 \times 5}{100}=0.25 \mathrm{W} \\ &\qquad\left(V \operatorname{across} R_{L}=5 \mathrm{V}\right) \end{aligned}
\begin{aligned} R_{L}&= R_{s}=100 \Omega \\ \therefore \quad P_{\max }&=\frac{V^{2}}{R}=\frac{5 \times 5}{100}=0.25 \mathrm{W} \\ &\qquad\left(V \operatorname{across} R_{L}=5 \mathrm{V}\right) \end{aligned}
There are 10 questions to complete.